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Proof step 2 from proof min-ins-01 (previous step)

a ∈ s, s.min ∈ s, s.max ∈ s ⊦ s = ∅, (s ++ b).min = s.min, # s = 0, a < s.min, s.max < a, b < a, b < s.min

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
a ∈ s, ∀ b. b ∈ s → ¬ b < a ⊦ a = s.min

Rule argument

spec: oset inst: name: min-q seq: a ∈ s, ∀ b. b ∈ s → ¬ b < a ⊦ a = s.min subst: [a, s] → [s.min, s ++ b] flags: rewrite, current

Additional information

Used simplifier rules: Max-max: ⊦ a ∈ s → ¬ s.max < a;%Min-min: ⊦ a ∈ s → ¬ a < s.min;%in: ⊦ a ∈ s → a ∈ s ++ b; from specification set-basic%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s); from specification set-basic Used terms: s.min, s ++ b